Difference between revisions of "2024 AMC 8 Problems/Problem 7"

Latest revision as of 06:29, 15 November 2024

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Video Solution by Math-X (First fully understand the problem!!!)
6 Video Solution (A Clever Explanation You’ll Get Instantly)
7 Video Solution 1 (easy to digest) by Power Solve
8 Video Solution by NiuniuMaths (Easy to understand!)
9 Video Solution 2 by SpreadTheMathLove
10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
11 Video Solution by Interstigation
12 Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
13 Video Solution by WhyMath
14 See Also

Problem

A $3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ , $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?

$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution 1

We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{\textbf{(E)\ 5}}$ .

Solution 2

Let $x$ be the number of $1$ by $1$ tiles. There are $21$ squares and each $2$ by $2$ or $1$ by $4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2$ by $2$ and $1$ by $4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1$ by $1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\boxed{\textbf{(E)\ 5}}$ , which can easily be confirmed to work.

~arfekete

Solution 3

Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have $4a + 4b + c = 21.$ Reducing modulo $4$ gives $c\equiv 1\pmod{4}$ , or $c = 1$ or $c = 5$ .

If $c = 1$ , then $a + b = 5$ . After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\boxed{\textbf{(E)\ 5}}$ , which can be constructed in many ways.

-Benedict T (countmath1)

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=WogR3xd8zMwJgzRf&t=712

~hsnacademy

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=L4ouVVVkFo4

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=578

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by WhyMath

https://youtu.be/LCCDCb1EHg0

@@ Line 23: / Line 23: @@
 -Benedict T (countmath1)
+==Video Solution by Math-X (First fully understand the problem!!!)==
+https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497
+~Math-X
 ==Video Solution (A Clever Explanation You’ll Get Instantly)==
@@ Line 32: / Line 37: @@
 https://youtu.be/16YYti_pDUg?si=KjRhUdCOAx10kgiW&t=59
-==Video Solution by Math-X (First fully understand the problem!!!)==
-https://youtu.be/BaE00H2SHQM?si=eg8T8wApi2j83Ia_&t=1497
-~Math-X
 ==Video Solution by NiuniuMaths (Easy to understand!)==

2024 AMC 8 (Problems • Answer Key • Resources)
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